A371673 Expansion of g.f. A(x) satisfying [x^(n-1)] A(x)^(n^2) = A000108(n-1) * n^n for n >= 1, where A000108 is the Catalan numbers.

1, 1, 2, 15, 284, 8575, 345460, 17190684, 1012901520, 68810750943, 5291667341342, 454479660308531, 43140290728900554, 4487833959824527910, 508072065566891421336, 62222074620010689986918, 8200304581300850453687880, 1157674985567876068399895997, 174357014524193551292388873190

Offset

0,3

Comments

Conjecture: a(n) is odd for n > 0 iff n = 2*A003714(k) + 1 for some k, where A003714 is the Fibbinary numbers (integers whose binary representation contains no consecutive ones). See A263075, A263190, and A171791.

Links

Paul D. Hanna, Table of n, a(n) for n = 0..300

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) [x^(n-1)] A(x)^(n^2) = n^n * binomial(2*n-1,n-1)/(2*n-1) for n >= 1.

(2) [x^(n-1)] A(x)^(n^2) = [x^(n-1)] 1/(1 - n*x)^n for n >= 1.

Example

G.f.: A(x) = 1 + x + 2*x^2 + 15*x^3 + 284*x^4 + 8575*x^5 + 345460*x^6 + 17190684*x^7 + 1012901520*x^8 + 68810750943*x^9 + 5291667341342*x^10 + ...
The table of coefficients of x^k in A(x)^(n^2) begin:
 n=1: [1,  1,    2,    15,    284,    8575,    345460, ...];
 n=2: [1,  4,   14,    88,   1365,   38304,   1497150, ...];
 n=3: [1,  9,   54,   363,   4410,  105705,   3874824, ...];
 n=4: [1, 16,  152,  1280,  13804,  263408,   8535648, ...];
 n=5: [1, 25,  350,  3875,  43750,  688205,  18352800, ...];
 n=6: [1, 36,  702, 10200, 133389, 1959552,  42189822, ...];
 n=7: [1, 49, 1274, 23863, 376320, 5810763, 108707676, ...];
 ...
where the terms along the main diagonal start as
 [1, 4, 54, 1280, 43750, 1959552, 108707676, ...]
which equals A000108(n-1)*n^n for n >= 1:
 [1, 1*2^2, 2*3^3, 5*4^4, 14*5^5, 42*6^6, 132*7^7, ...].
Compare the above table to the coefficients in 1/(1 - n*x)^n:
 n=1: [1,  1,    1,     1,      1,       1,         1, ...];
 n=2: [1,  4,   12,    32,     80,     192,       448, ...];
 n=3: [1,  9,   54,   270,   1215,    5103,     20412, ...];
 n=4: [1, 16,  160,  1280,   8960,   57344,    344064, ...];
 n=5: [1, 25,  375,  4375,  43750,  393750,   3281250, ...];
 n=6: [1, 36,  756, 12096, 163296, 1959552,  21555072, ...];
 n=7: [1, 49, 1372, 28812, 504210, 7764834, 108707676, ...];
 ...
to see that the main diagonals are equal.

Prog

(PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A;
A[m] = ( m^m*binomial(2*m-1, m-1)/(2*m-1) - Vec( Ser(A)^(m^2) )[m] )/(m^2) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A263075, A263190, A171791.

Cf. A000108, A003714, A118113.

Sequence in context: A143886 A174482 A076111 * A087526 A283275 A372312

Adjacent sequences: A371670 A371671 A371672 * A371674 A371675 A371676

Keyword

nonn

Author

Paul D. Hanna, Apr 02 2024

Status

approved